Norm inequality for sum and difference of positive-definite matrices If $X_{1}$ and $X_{2}$ are positive definite matrices, how to show
that $\left\Vert X_{1}-X_{2}\right\Vert \le\left\Vert X_{1}+X_{2}\right\Vert$ for the spectral norm? and how about for the nuclear norm? Thanks in advance.
 A: This answer only answers the first question.  It does so in a way that may be unnecessarily complicated for your taste.

If $A$ and $B$ are Hermitian matrices, let $A\leq B$ mean that $B-A$ is positive semidefinite.  A Hermitian matrix is positive semidefinite if and only if all of its eigenvalues are nonnegative, and the spectral norm of a Hermitian matrix is the maximum of the absolute values of its eigenvalues.
Then $$-\|X_2\|I\leq -X_2\leq X_1-X_2\leq X_1\leq \|X_1\|I.$$  This implies that all of the eigenvalues of $X_1-X_2$ lie in the interval $[-\|X_2\|,\|X_1\|]$, which implies that $\|X_1-X_2\|\leq \max\{\|X_1\|,\|X_2\|\}$.
On the other hand, $0\leq X_1\leq X_1+X_2$ implies that $\|X_1\|\leq\|X_1+X_2\|$, and similarly $\|X_2\|\leq \|X_1+X_2\|$, so $\max\{\|X_1\|,\|X_2\|\}\leq \|X_1+X_2\|$.
A: As $X_1,X_2$ are positive semidefinite, Courant-Fischer minimax principle implies that $\lambda_i(X_1-X_2),\lambda_i(X_2-X_1)\le\lambda_i(X_1+X_2)$, where $\lambda_i$ denotes the $i$-th eigenvalue of a matrix when the spectrum is arranged in ascending or descending order. Hence we get $\rho(X_1-X_2)\le\rho(X_1+X_2)$ and $\sum_i|\lambda_i(X_1-X_2)|\le\sum_i\lambda_i(X_1+X_2)$, i.e. your inequality holds for both the spectral norm and the nuclear norm.
